A triangular number is a term from the sequence generated by this equation (in Perl):

foreach my $n (1..1000) { $x=.5*$n*($n+1); print $x; }

This gives 1, 3, 6, 10 and so on. This months challenge is to find a certain four numbers from this sequence. They don't give us the numbers, but they have substituted letters for digits of these numbers, and made words from them. The words are: ONE, THREE, SIX and TEN.

So for instance, we could have O=1, N=2, E=3, and the first number would be 123.

So I wrote a Perl program to have a go at it. I make a hash of the letters, assign a value (starting at zero) to each letter, then start incrementing. It's been going for the last few hours and I just realised it could take quite a long time to finish.

I am declaring that the challenge is to make it efficient, not write it with the fewest characters. This is not part of the official challenge, if you feel like writing it in one line, go right ahead.

The prize is drawn on the 28 June

**Update**

jeroenes asked if there were restrictions on the numbers - there are. It's one letter for one digit. The digits may be 0..9 (the article is a little vague on this, but that's my reading). **But I forgot to mention that none of the numbers start with a zero. i.e. 003 is not a valid solution for this challenge**

The challenge does not mention the order of the numbers, so I guess we can assume 'any'.

If you would also like to apply for the 15-pound prize you can send your solution to enigma@newscientist.com (include your postal address)

#!/usr/bin/perl use strict; my @words=('ONE', 'THREE', 'SIX', 'TEN'); my %letters; foreach (@words) { my @a=split //, $_; foreach (@a) { $letters{$_}=0; } } #Construct an array of the letters (you'll see why #in a few lines) my @letters=(keys %letters); my $count; my $oldtime; do { $letters{$letters[$#letters]}++; foreach( reverse(0..$#letters)) { if ($count++>10000) { print time-$oldtime,"\n"; $oldtime=time; $count=0; } #Increment the number for the letter at the end #of the array, and don't forget to carry if ($letters{$letters[$_]} > 9) { $letters{$letters[$_]}=0; $letters{$letters[$_-1]}++; } #Turn the letters into numbers prep(%letters); } } until ($letters{$letters[0]} > 9); sub prep { my %letters=@_; my $match=0; my %match; foreach my $a (@words) { my $test= $a; #Change the letters into numbers foreach (keys %letters) { $test=~ s/$_/$letters{$_}/g; } $match{"$a=$test is triangular\n"}=1 and $match ++ if is_triangular +($test); } if ($match>0) { # print "-----------\n"; if ($match>(@words-1)) { print $_ foreach keys %match; print "Found it!\n"; #exit; } } } #This should be memoised, but I'm afraid of the #memory blowout sub is_triangular { my $num=shift; my ($n,$x)=(1,0); do { $x=.5*$n*(++$n); return 1 if $x==$num; } until $x > $num; return 0; }

**Update II**

Since tachyon has indeed beaten me to the answer I shall enter PerlMonks in the competetion, winnings go in the plate. (More likely they will go into an envelope and off to vroom).

**Update III**

My bad. The challenge for the magazine is to get the answer. The answers are then drawn out of a hat, first right one gets the prize.

**My** challenge to you is to do it most efficiently.

____________________

Jeremy

I didn't believe in evil until I dated it.

Comment on(Efficiency Golf) Triangular numbersSelectorDownloadCode